Question

\[\int e^x \frac{x - 1}{\left( x + 1 \right)^3} \text{ dx }\]

Answer 1

\[\text{ Let I } = \int e^x \left[ \frac{x - 1}{\left( x - 1 \right)^3} \right]dx\]

\[ = \int e^x \left[ \frac{x + 1 - 2}{\left( x + 1 \right)^3} \right]dx\]

\[ = \int e^x \left[ \frac{1}{\left( x - 1 \right)^2} - \frac{2}{\left( x + 1 \right)^3} \right]dx\]

\[\text{ Here}, f(x) = \frac{1}{\left( x + 1 \right)^2}\]

\[ \Rightarrow f'(x) = \frac{- 2}{\left( x + 1 \right)^2}\]

\[\text{ Put e}^x f(x) = t\]

\[\text{  let e}^x \frac{1}{\left( x + 1 \right)^2} = t\]

\[\text{ Diff  both  sides }\]

\[ e^x \frac{1}{\left( x + 1 \right)^2} + e^x \frac{\left( - 2 \right)}{\left( x + 1 \right)^3} = \frac{dt}{dx}\]

\[ \Rightarrow e^x \left[ \frac{1}{\left( x + 1 \right)^2} - \frac{2}{\left( x + 1 \right)^3} \right]dx = dt\]

\[ \therefore \int e^x \left[ \frac{1}{\left( x + 1 \right)^2} - \frac{2}{\left( x + 1 \right)^3} \right]dx = \int dt\]

\[ = t + C\]

\[ = \frac{e^x}{\left( x + 1 \right)^2} + C\]